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[Leetcode]232. Implement Queue using Stacks(C++)

题目描述

题目链接:232. Implement Queue using Stacks

Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

  • void push(int x) Pushes element x to the back of the queue.
  • int pop() Removes the element from the front of the queue and returns it.
  • int peek() Returns the element at the front of the queue.
  • boolean empty() Returns true if the queue is empty, false otherwise.

例子

例子 1

Input:
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output: [null, null, null, 1, 1, false]
Explanation:

Follow Up

Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

Note

You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack’s standard operations.

解题思路

本题要求用栈实现队列的功能,这两个数据结构的区别在于数据的出入顺序不同,栈是先入后出(FILO),队列是先入先出(FIFO)。因此实现的关键的是逆转栈中存储数据的顺序,题目提示可以用两个栈,因此只要在每次 push 的时候将维护的栈的元素弹出到另一个栈中,压入新元素,将重新将新栈的元素压回原栈中,此时栈顶元素即为队列首元素。代码如下:

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#include <stack>

class MyQueue {
public:
/** Initialize your data structure here. */
MyQueue() {}

/** Push element x to the back of queue. */
void push(int x) {
std::stack<int> temp = ordered_;
std::stack<int> reversed;
while (!temp.empty()) {
reversed.push(temp.top());
temp.pop();
}
reversed.push(x);
temp.swap(reversed);
ordered_ = std::stack<int>();
while (!temp.empty()) {
ordered_.push(temp.top());
temp.pop();
}
}

/** Removes the element from in front of queue and returns that element. */
int pop() {
int val = ordered_.top();
ordered_.pop();
return val;
}

/** Get the front element. */
int peek() { return ordered_.top(); }

/** Returns whether the queue is empty. */
bool empty() { return ordered_.empty(); }

private:
std::stack<int> ordered_;
};
  • 时间复杂度:
    • push: O(n)
    • pop: O(1)
    • peek: O(1)
    • empty: O(1)
  • 空间复杂度: O(n)

GitHub 代码同步地址: 232.ImplementQueueUsingStacks.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions